Mathematics: Trignometry Trignometricratioqudrant Mock Test – Free Online Practice

Mathematics: Trignometry Trignometricratioqudrant ke liye free mock test: 28 questions, timer, negative marking aur explanation ke saath. Koi login nahi, koi payment nahi — turant shuru karein.

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Practice Questions (28 of 28)

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1. If cos x = - 3/5, x lies in the third quadrant, find the values of other five trigonometric functions.
   यदि cos x = - 3/5 है और x तीसरे चतुर्थांश में स्थित है, तो अन्य पांच त्रिकोणमितीय फलनों के मान ज्ञात कीजिए।
   (A) (a) sin x = -4/5, tan x = 4/3, cot x = 3/4, sec x = -5/3, cosec x = -5/4

   (B) (b) sin x = 4/5, tan x = -4/3, cot x = -3/4, sec x = -5/3, cosec x = 5/4

   (C) (c) sin x = -4/5, tan x = -4/3, cot x = -3/4, sec x = 5/3, cosec x = -5/4

   (D) (d) sin x = 3/5, tan x = 4/3, cot x = 3/4, sec x = -5/3, cosec x = 5/3
   ✅ Answer & Explanation

   Sahi jawab: A) (a) sin x = -4/5, tan x = 4/3, cot x = 3/4, sec x = -5/3, cosec x = -5/4

   Explanation: Method 1 (Quadrant Rule & Triplet Method): Step 1: ASTC rule (All-Sin-Tan-Cos) ke mutabiq 3rd quadrant me keval $\tan$ aur $\cot$ positive hote hain, baaki sabhi ($\sin, \cos, \sec, \csc$) negative hote hain. Step 2: Triplet $3-4-5$ se: Base $= 3$, Hypotenuse $= 5 \implies$ Perpendicular $= 4$. Step 3: Ratios determine karein: - $\sin x = -\frac{P}{H} = -\frac{4}{5}$ - $\tan x = +\frac{P}{B} = \frac{4}{3}$ - $\cot x = +\frac{B}{P} = \frac{3}{4}$ - $\sec x = -\frac{H}{B} = -\frac{5}{3}$ - $\csc x = -\frac{H}{P} = -\frac{5}{4}$ Correct option (a) hai. Method 2 (Identity Method): $$\sin x = -\sqrt{1 - \cos^2 x} = -\sqrt{1 - \frac{9}{25}} = -\frac{4}{5}$$ $$\tan x = \frac{\sin x}{\cos x} = \frac{-4/5}{-3/5} = \frac{4}{3}, \quad \cot x = \frac{3}{4}, \quad \sec x = -\frac{5}{3}, \quad \csc x = -\frac{5}{4}$$ Sahi uttar (a) hai.
2. If cos x = - 5/13, x lies in the second quadrant, find the values of other five trigonometric functions.
   यदि cos x = - 5/13 है और x दूसरे चतुर्थांश में स्थित है, तो अन्य पांच त्रिकोणमितीय फलनों के मान ज्ञात कीजिए।
   (A) (a) sin x = 12/13, tan x = -12/5, cot x = -5/12, sec x = -13/5, cosec x = 13/12

   (B) (b) sin x = -12/13, tan x = 12/5, cot x = 5/12, sec x = -13/5, cosec x = -13/12

   (C) (c) sin x = 12/13, tan x = 12/5, cot x = 5/12, sec x = 13/5, cosec x = 13/12

   (D) (d) sin x = -12/13, tan x = -12/5, cot x = -5/12, sec x = -13/5, cosec x = 13/12
   ✅ Answer & Explanation

   Sahi jawab: A) (a) sin x = 12/13, tan x = -12/5, cot x = -5/12, sec x = -13/5, cosec x = 13/12

   Explanation: Method 1 (Quadrant Rule & Triplet Method): Step 1: 2nd quadrant me keval $\sin$ aur $\csc$ positive hote hain, baaki sabhi negative hote hain. Step 2: Triplet $5-12-13$ se: Base $= 5$, Hypotenuse $= 13 \implies$ Perpendicular $= 12$. Step 3: Signs apply karte hue values likhein: - $\sin x = +\frac{12}{13}$ - $\csc x = +\frac{13}{12}$ - $\tan x = -\frac{12}{5}$ - $\cot x = -\frac{5}{12}$ - $\sec x = -\frac{13}{5}$ Correct option (a) hai. Method 2 (Identity Method): $$\sin x = +\sqrt{1 - \cos^2 x} = \sqrt{1 - \frac{25}{169}} = \frac{12}{13}$$ $$\tan x = \frac{\sin x}{\cos x} = \frac{12/13}{-5/13} = -\frac{12}{5}$$ Sahi uttar (a) hai.
3. The value of sin 960° cos 330° + cos 120° sin 150° is :
   sin 960° cos 330° + cos 120° sin 150° का मान क्या होगा?
   (A) (a) - 1

   (B) (b) 1

   (C) (c) 1/√2

   (D) (d) √3/2
   ✅ Answer & Explanation

   Sahi jawab: A) (a) - 1

   Explanation: Method 1 (Allied Angles Conversion Method): Step 1: Sabhi angles ko standard form me reduce karein: - $\sin 960^\circ = \sin(2 \times 360^\circ + 240^\circ) = \sin 240^\circ = \sin(180^\circ + 60^\circ) = -\sin 60^\circ = -\frac{\sqrt{3}}{2}$ - $\cos 330^\circ = \cos(360^\circ - 30^\circ) = +\cos 30^\circ = \frac{\sqrt{3}}{2}$ - $\cos 120^\circ = \cos(180^\circ - 60^\circ) = -\cos 60^\circ = -\frac{1}{2}$ - $\sin 150^\circ = \sin(180^\circ - 30^\circ) = +\sin 30^\circ = \frac{1}{2}$ Step 2: Values substitute karein: $$\text{Value} = \left(-\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)$$ $$\text{Value} = -\frac{3}{4} - \frac{1}{4} = -\frac{4}{4} = -1$$ Correct option (a) - 1 hai. Method 2 (Compound Angle Identity Method): $$\sin 960^\circ\cos 330^\circ + \cos 120^\circ\sin 150^\circ = (-\sin 60^\circ)(\cos 30^\circ) + (-\cos 60^\circ)(\sin 30^\circ)$$ $$= -(\sin 60^\circ\cos 30^\circ + \cos 60^\circ\sin 30^\circ) = -\sin(60^\circ + 30^\circ) = -\sin 90^\circ = -1$$ Sahi uttar (a) hai.
4. Find the value of sin(31π/3).
   sin(31π/3) का मान ज्ञात कीजिए।
   (A) (a) 1

   (B) (b) 1/2

   (C) (c) √3/2

   (D) (d) 0
   ✅ Answer & Explanation

   Sahi jawab: C) (c) √3/2

   Explanation: Method 1 (2nπ Periodicity Method): Step 1: Fraction $\frac{31\pi}{3}$ ko $2k\pi + \theta$ ke form me break karein: $$\frac{31\pi}{3} = 10\pi + \frac{\pi}{3} = 5(2\pi) + \frac{\pi}{3}$$ Step 2: Periodicity property: $\sin(2k\pi + \theta) = \sin\theta$: $$\sin\left(\frac{31\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right)$$ Step 3: $\sin(\pi/3) = \sin 60^\circ = \frac{\sqrt{3}}{2}$: Correct option (c) √3/2 hai. Method 2 (Degree Conversion Method): $$\frac{31 \times 180^\circ}{3} = 31 \times 60^\circ = 1860^\circ$$ $$1860^\circ = 5 \times 360^\circ + 60^\circ \implies \sin 1860^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}$$ Sahi uttar (c) hai.
5. Find the value of cos(- 1710°).
   cos(- 1710°) का मान ज्ञात कीजिए।
   (A) (a) 0

   (B) (b) 1

   (C) (c) - 1

   (D) (d) 1/2
   ✅ Answer & Explanation

   Sahi jawab: A) (a) 0

   Explanation: Method 1 (Negative Angle & Multiple of 360° Method): Step 1: Property: $\cos(-\theta) = \cos\theta$: $$\cos(-1710^\circ) = \cos(1710^\circ)$$ Step 2: 1710° ko $360^\circ$ ke multiples ke roop me express karein ($360^\circ \times 5 = 1800^\circ$): $$1710^\circ = 1800^\circ - 90^\circ = 5(360^\circ) - 90^\circ$$ Step 3: Evaluate karein: $$\cos(5 \times 360^\circ - 90^\circ) = \cos(-90^\circ) = \cos(90^\circ) = 0$$ Correct option (a) 0 hai. Method 2 (Quotient Step Method): $$1710^\circ = 4 \times 360^\circ + 270^\circ$$ $$\cos(1710^\circ) = \cos(270^\circ) = 0$$ Sahi uttar (a) hai.
6. Solve tan(13π/12).
   tan(13π/12) का मान क्या होगा?
   (A) (a) 2 + √3

   (B) (b) 2 - √3

   (C) (c) √3

   (D) (d) √3 + 1
   ✅ Answer & Explanation

   Sahi jawab: B) (b) 2 - √3

   Explanation: Method 1 (Allied Angle & 15° Standard Value Method): Step 1: Angle ko split karein: $$\frac{13\pi}{12} = \pi + \frac{\pi}{12}$$ Step 2: 3rd quadrant me tangent positive hota hai ($\tan(\pi + \theta) = \tan\theta$): $$\tan\left(\frac{13\pi}{12}\right) = \tan\left(\frac{\pi}{12}\right) = \tan 15^\circ$$ Step 3: Standard 15° formula yaad rakhein: $$\tan 15^\circ = 2 - \sqrt{3}$$ Correct option (b) 2 - √3 hai. Method 2 (Compound Angle tan(45° - 30°) Method): $$\tan 15^\circ = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ\tan 30^\circ} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3}$$ Sahi uttar (b) hai.
7. Find 3 sin(π/6) sec(π/3) - 4 sin(5π/6) cot(π/4) ?
   3 sin(π/6) sec(π/3) - 4 sin(5π/6) cot(π/4) का मान ज्ञात कीजिए?
   (A) (a) 1

   (B) (b) 0

   (C) (c) - 1

   (D) (d) 2
   ✅ Answer & Explanation

   Sahi jawab: A) (a) 1

   Explanation: Method 1 (Step-by-Step Direct Value Substitution Method): Step 1: Pratyek term ki value nikalte hain: - $\sin(\pi/6) = \sin 30^\circ = \frac{1}{2}$ - $\sec(\pi/3) = \sec 60^\circ = 2$ - $\sin(5\pi/6) = \sin(\pi - \pi/6) = \sin(\pi/6) = \frac{1}{2}$ - $\cot(\pi/4) = \cot 45^\circ = 1$ Step 2: Expression me values rakhein: $$\text{Term 1} = 3 \times \left(\frac{1}{2}\right) \times 2 = 3$$ $$\text{Term 2} = 4 \times \left(\frac{1}{2}\right) \times 1 = 2$$ Step 3: Subtract karein: $$\text{Value} = 3 - 2 = 1$$ Correct option (a) 1 hai. Method 2 (Inspection Method): $$3\left(\frac{1}{2}\right)(2) - 4\left(\frac{1}{2}\right)(1) = 3 - 2 = 1$$ Sahi uttar (a) hai.
8. Find sin²(π/6) + cos²(π/3) - tan²(π/4) = ?
   sin²(π/6) + cos²(π/3) - tan²(π/4) का मान ज्ञात कीजिए?
   (A) (a) 1/2

   (B) (b) - 1/2

   (C) (c) 1

   (D) (d) - 1
   ✅ Answer & Explanation

   Sahi jawab: B) (b) - 1/2

   Explanation: Method 1 (Standard Angles Table Method): Step 1: Standard values put karein: - $\sin(\pi/6) = \frac{1}{2} \implies \sin^2(\pi/6) = \frac{1}{4}$ - $\cos(\pi/3) = \frac{1}{2} \implies \cos^2(\pi/3) = \frac{1}{4}$ - $\tan(\pi/4) = 1 \implies \tan^2(\pi/4) = 1$ Step 2: Expression evaluate karein: $$\frac{1}{4} + \frac{1}{4} - 1 = \frac{1}{2} - 1 = -\frac{1}{2}$$ Correct option (b) - 1/2 hai. Method 2 (Direct Fraction Calculation): $$0.25 + 0.25 - 1 = 0.5 - 1 = -0.5 = -\frac{1}{2}$$ Sahi uttar (b) hai.
9. Find 2 sin²(3π/4) + 2 cos²(π/4) + 2 sec²(π/3) = ?
   2 sin²(3π/4) + 2 cos²(π/4) + 2 sec²(π/3) का मान क्या होगा?
   (A) (a) 4

   (B) (b) 5

   (C) (c) 10

   (D) (d) 6
   ✅ Answer & Explanation

   Sahi jawab: C) (c) 10

   Explanation: Method 1 (Standard Radians Evaluation Method): Step 1: Har term ko evaluate karein: - $\sin(3\pi/4) = \sin(\pi - \pi/4) = \sin(\pi/4) = \frac{1}{\sqrt{2}} \implies 2\sin^2(3\pi/4) = 2\left(\frac{1}{2}\right) = 1$ - $\cos(\pi/4) = \frac{1}{\sqrt{2}} \implies 2\cos^2(\pi/4) = 2\left(\frac{1}{2}\right) = 1$ - $\sec(\pi/3) = 2 \implies 2\sec^2(\pi/3) = 2(2^2) = 2(4) = 8$ Step 2: Teeno terms ko add karein: $$\text{Total} = 1 + 1 + 8 = 10$$ Correct option (c) 10 hai. Method 2 (Direct Summation): $$2\left(\frac{1}{2}\right) + 2\left(\frac{1}{2}\right) + 2(4) = 1 + 1 + 8 = 10$$ Sahi uttar (c) hai.
10. Find 2 sin²(π/6) + cosec²(7π/6) cos²(π/3) = ?
    2 sin²(π/6) + cosec²(7π/6) cos²(π/3) का मान क्या होगा?
    (A) (a) 1

    (B) (b) 1/2

    (C) (c) - 1/2

    (D) (d) 3/2
    ✅ Answer & Explanation

    Sahi jawab: D) (d) 3/2

    Explanation: Method 1 (Allied Angles & Square Method): Step 1: First term evaluate karein: $$2\sin^2\left(\frac{\pi}{6}\right) = 2\left(\frac{1}{2}\right)^2 = 2\left(\frac{1}{4}\right) = \frac{1}{2}$$ Step 2: Second term evaluate karein: $$\frac{7\pi}{6} = \pi + \frac{\pi}{6} \implies \csc\left(\frac{7\pi}{6}\right) = -\csc\left(\frac{\pi}{6}\right) = -2$$ $$\csc^2\left(\frac{7\pi}{6}\right) = (-2)^2 = 4$$ $$\cos^2\left(\frac{\pi}{3}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$ $$\text{Second Term} = 4 \times \frac{1}{4} = 1$$ Step 3: Dono ko add karein: $$\text{Value} = \frac{1}{2} + 1 = \frac{3}{2}$$ Correct option (d) 3/2 hai. Method 2 (Direct Addition): $$\frac{1}{2} + 1 = 1.5 = \frac{3}{2}$$ Sahi uttar (d) hai.
11. Find cot²(π/6) + cosec(5π/6) + 3 tan²(π/6) = ?
    cot²(π/6) + cosec(5π/6) + 3 tan²(π/6) का मान क्या होगा?
    (A) (a) 6

    (B) (b) 4

    (C) (c) 5

    (D) (d) 3
    ✅ Answer & Explanation

    Sahi jawab: A) (a) 6

    Explanation: Method 1 (Step-by-Step Standard Substitution Method): Step 1: Pratyek pad ka man nikalte hain: - $\cot(\pi/6) = \cot 30^\circ = \sqrt{3} \implies \cot^2(\pi/6) = (\sqrt{3})^2 = 3$ - $\csc(5\pi/6) = \csc(\pi - \pi/6) = \csc(\pi/6) = 2$ - $\tan(\pi/6) = \tan 30^\circ = \frac{1}{\sqrt{3}} \implies 3\tan^2(\pi/6) = 3\left(\frac{1}{3}\right) = 1$ Step 2: Teeno ko add karein: $$\text{Total} = 3 + 2 + 1 = 6$$ Correct option (a) 6 hai. Method 2 (Direct Inspection): $$3 + 2 + 1 = 6$$ Sahi uttar (a) hai.
12. Find [cos(π + x) cos(-x)] / [sin(π - x) cos(π/2 + x)] = ?
    [cos(π + x) cos(-x)] / [sin(π - x) cos(π/2 + x)] का मान क्या होगा?
    (A) (a) sin²x

    (B) (b) tan²x

    (C) (c) - cot²x

    (D) (d) cot²x
    ✅ Answer & Explanation

    Sahi jawab: D) (d) cot²x

    Explanation: Method 1 (Allied Angles Quadrant Formulas Method): Step 1: Har trigonometric term ko simplify karein: - $\cos(\pi + x) = -\cos x$ (3rd quadrant me cosine negative hota hai) - $\cos(-x) = \cos x$ (Even function property) - $\sin(\pi - x) = \sin x$ (2nd quadrant me sine positive hota hai) - $\cos(\pi/2 + x) = -\sin x$ (2nd quadrant me change hokar negative sine banta hai) Step 2: Numerator aur Denominator multiply karein: $$\text{Numerator} = (-\cos x)(\cos x) = -\cos^2 x$$ $$\text{Denominator} = (\sin x)(-\sin x) = -\sin^2 x$$ Step 3: Fraction solve karein: $$\frac{-\cos^2 x}{-\sin^2 x} = \frac{\cos^2 x}{\sin^2 x} = \cot^2 x$$ Correct option (d) cot²x hai. Method 2 (Value-Putting Method): $x = 45^\circ$ assume karein: $$\frac{\cos 225^\circ \cos(-45^\circ)}{\sin 135^\circ \cos 135^\circ} = \frac{(-1/\sqrt{2})(1/\sqrt{2})}{(1/\sqrt{2})(-1/\sqrt{2})} = 1$$ $\cot^2 45^\circ = 1$ (Option d matches). Sahi uttar (d) hai.
13. sin 15° + cos 105° =
    sin 15° + cos 105° का मान किसके बराबर है?
    (A) (a) 0

    (B) (b) 2 sin 15°

    (C) (c) cos 15° + sin 15°

    (D) (d) 2 cos 15°
    ✅ Answer & Explanation

    Sahi jawab: A) (a) 0

    Explanation: Method 1 (Complementary & Allied Angles Method): Step 1: $\cos 105^\circ$ ko convert karein: $$\cos 105^\circ = \cos(90^\circ + 15^\circ) = -\sin 15^\circ$$ Step 2: Expression me substitute karein: $$\sin 15^\circ + \cos 105^\circ = \sin 15^\circ + (-\sin 15^\circ) = \sin 15^\circ - \sin 15^\circ = 0$$ Correct option (a) 0 hai. Method 2 (cos C + cos D Transformation Method): $\sin 15^\circ = \cos 75^\circ$. $$\cos 75^\circ + \cos 105^\circ = 2\cos\left(\frac{75^\circ + 105^\circ}{2}\right)\cos\left(\frac{105^\circ - 75^\circ}{2}\right) = 2\cos 90^\circ\cos 15^\circ$$ Chuki $\cos 90^\circ = 0$ hota hai, isliye $2(0)\cos 15^\circ = 0$. Sahi uttar (a) hai.
14. cos 25° + cos 5° + cos 175° + cos 205° + cos 300° = ?
    cos 25° + cos 5° + cos 175° + cos 205° + cos 300° का मान क्या होगा?
    (A) (a) 1/2

    (B) (b) - 1/2

    (C) (c) √3/2

    (D) (d) 1
    ✅ Answer & Explanation

    Sahi jawab: A) (a) 1/2

    Explanation: Method 1 (Allied Angles Cancellation Method): Step 1: Angles ke pairs banate hain jo $180^\circ$ se related hain: - $\cos 175^\circ = \cos(180^\circ - 5^\circ) = -\cos 5^\circ \implies \cos 5^\circ + \cos 175^\circ = 0$ - $\cos 205^\circ = \cos(180^\circ + 25^\circ) = -\cos 25^\circ \implies \cos 25^\circ + \cos 205^\circ = 0$ Step 2: Dono pairs cancel hone ke baad keval aakhiri term $\cos 300^\circ$ bachta hai: $$\text{Remaining} = \cos 300^\circ$$ Step 3: $\cos 300^\circ$ ki value evaluate karein: $$\cos 300^\circ = \cos(360^\circ - 60^\circ) = \cos 60^\circ = \frac{1}{2}$$ Correct option (a) 1/2 hai. Method 2 (cos A + cos(180° ± A) Identity Method): Formula: $\cos A + \cos(180^\circ - A) = 0$ aur $\cos A + \cos(180^\circ + A) = 0$. Isliye pehle chaar terms ka sum 0 ho jayega: $$0 + 0 + \cos(360^\circ - 60^\circ) = \cos 60^\circ = \frac{1}{2}$$ Sahi uttar (a) hai.
15. What is the value of [cos(π/2 - 3A) - cos(π/2 + A)] / [cos A + cos(π + 3A)] ?
    [cos(π/2 - 3A) - cos(π/2 + A)] / [cos A + cos(π + 3A)] का मान क्या होगा?
    (A) (a) tan 2A

    (B) (b) cot A

    (C) (c) tan 2A

    (D) (d) cot 2A
    ✅ Answer & Explanation

    Sahi jawab: B) (b) cot A

    Explanation: Method 1 (Allied Angles & CD Formula Method): Step 1: Allied angle properties lagakar simplify karein: - $\cos(\pi/2 - 3A) = \sin 3A$ - $\cos(\pi/2 + A) = -\sin A$ - $\cos(\pi + 3A) = -\cos 3A$ Step 2: Numerator aur Denominator me values substitute karein: $$\text{Numerator} = \sin 3A - (-\sin A) = \sin 3A + \sin A$$ $$\text{Denominator} = \cos A - \cos 3A$$ Step 3: CD formulas apply karein: $$\sin 3A + \sin A = 2\sin 2A\cos A$$ $$\cos A - \cos 3A = 2\sin 2A\sin A$$ Step 4: Divide karein: $$\frac{2\sin 2A\cos A}{2\sin 2A\sin A} = \frac{\cos A}{\sin A} = \cot A$$ Correct option (b) cot A hai. Method 2 (Value-Putting Method): Angle $A = 30^\circ$ assume karein: $$\text{Numerator} = \cos(0^\circ) - \cos(120^\circ) = 1 - (-1/2) = \frac{3}{2}$$ $$\text{Denominator} = \cos 30^\circ + \cos(270^\circ) = \frac{\sqrt{3}}{2} + 0 = \frac{\sqrt{3}}{2}$$ $$\text{Value} = \frac{3/2}{\sqrt{3}/2} = \sqrt{3}$$ $\cot 30^\circ = \sqrt{3}$ hota hai, jo option (b) se match karta hai. Sahi uttar (b) hai.
16. If ABCD is a cyclic quadrilateral then cosA + cosB + cosC + cosD = ?
    यदि ABCD एक चक्रीय चतुर्भुज (cyclic quadrilateral) है, तो cosA + cosB + cosC + cosD का मान क्या होगा?
    (A) (a) 2(cosA + cosC)

    (B) (b) 2(cosD + cosB)

    (C) (c) 2(sinA + sinC)

    (D) (d) 0
    ✅ Answer & Explanation

    Sahi jawab: D) (d) 0

    Explanation: Method 1 (Cyclic Quadrilateral Property Method): Step 1: Cyclic quadrilateral ke sammukh (opposite) konon ka yog $180^\circ$ hota hai: $$A + C = 180^\circ \implies C = 180^\circ - A$$ $$B + D = 180^\circ \implies D = 180^\circ - B$$ Step 2: Cosine apply karein: $$\cos C = \cos(180^\circ - A) = -\cos A$$ $$\cos D = \cos(180^\circ - B) = -\cos B$$ Step 3: Expression me substitute karein: $$\cos A + \cos B + (-\cos A) + (-\cos B) = 0$$ Correct option (d) 0 hai. Method 2 (Value-Putting Method): ABCD ko rectangle assume karein (har angle $90^\circ$): $$\cos 90^\circ + \cos 90^\circ + \cos 90^\circ + \cos 90^\circ = 0 + 0 + 0 + 0 = 0$$ Sahi uttar (d) hai.
17. tan θ · sin(π/2 + θ) · cos(π/2 - θ) = ?
    tan θ · sin(π/2 + θ) · cos(π/2 - θ) का मान क्या होगा?
    (A) (a) 1

    (B) (b) 0

    (C) (c) cos θ

    (D) (d) sin²θ
    ✅ Answer & Explanation

    Sahi jawab: D) (d) sin²θ

    Explanation: Method 1 (Allied Angles Transformation Method): Step 1: Quadrant rules lagakar trigonometric ratios convert karein: - $\sin(\pi/2 + \theta) = \cos\theta$ - $\cos(\pi/2 - \theta) = \sin\theta$ Step 2: Expression me multiply karein: $$\tan\theta \cdot \cos\theta \cdot \sin\theta$$ Step 3: $\tan\theta = \frac{\sin\theta}{\cos\theta}$ substitute karein: $$\frac{\sin\theta}{\cos\theta} \cdot \cos\theta \cdot \sin\theta = \sin^2\theta$$ Correct option (d) sin²θ hai. Method 2 (Value-Putting Method): Angle $\theta = 45^\circ$ assume karein: $$\tan 45^\circ \cdot \sin 135^\circ \cdot \cos 45^\circ = 1 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{1}{2}$$ $\sin^2 45^\circ = (1/\sqrt{2})^2 = 1/2$ (Verified). Sahi uttar (d) hai.
18. tan A + tan(180° + A) + cot(90° + A) + cot(360° - A) = ?
    tan A + tan(180° + A) + cot(90° + A) + cot(360° - A) का मान क्या होगा?
    (A) (a) 0

    (B) (b) 2 tanA

    (C) (c) 2 cotA

    (D) (d) tanA - cotA
    ✅ Answer & Explanation

    Sahi jawab: D) (d) tanA - cotA

    Explanation: Method 1 (ASTC Quadrant Rule Method): Step 1: Pratyek pad ko quadrant rules se simplify karein: - $\tan(180^\circ + A) = +\tan A$ (3rd quadrant me tangent positive hota hai) - $\cot(90^\circ + A) = -\tan A$ (2nd quadrant me cotangent negative hota hai) - $\cot(360^\circ - A) = -\cot A$ (4th quadrant me cotangent negative hota hai) Step 2: Sabhi values ko add karein: $$\tan A + \tan A + (-\tan A) + (-\cot A) = 2\tan A - \tan A - \cot A = \tan A - \cot A$$ Correct option (d) tanA - cotA hai. Method 2 (Value-Putting Method): $A = 30^\circ$ assume karein: $$\tan 30^\circ + \tan 210^\circ + \cot 120^\circ + \cot 330^\circ$$ $$= \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{3}} - \sqrt{3} = \frac{1}{\sqrt{3}} - \sqrt{3} = \tan 30^\circ - \cot 30^\circ$$ Sahi uttar (d) hai.
19. cot(π/7) + cot(2π/7) + cot(3π/7) + .........+ cot(6π/7) = ?
    cot(π/7) + cot(2π/7) + cot(3π/7) + .........+ cot(6π/7) का मान क्या होगा?
    (A) (a) 0

    (B) (b) 1

    (C) (c) - 1

    (D) (d) 2
    ✅ Answer & Explanation

    Sahi jawab: A) (a) 0

    Explanation: Method 1 (Symmetric Pair Cancellation Method): Step 1: Formula: $\cot(\pi - \theta) = -\cot\theta$. Step 2: Shuru aur aakhiri terms ke pairs banate hain: - $\cot(6\pi/7) = \cot(\pi - \pi/7) = -\cot(\pi/7) \implies \cot(\pi/7) + \cot(6\pi/7) = 0$ - $\cot(5\pi/7) = \cot(\pi - 2\pi/7) = -\cot(2\pi/7) \implies \cot(2\pi/7) + \cot(5\pi/7) = 0$ - $\cot(4\pi/7) = \cot(\pi - 3\pi/7) = -\cot(3\pi/7) \implies \cot(3\pi/7) + \cot(4\pi/7) = 0$ Step 3: Sabhi teen pairs ka sum zero ho jata hai: $$\text{Total} = 0 + 0 + 0 = 0$$ Correct option (a) 0 hai. Method 2 (Direct Property Inspection): Jab angles ka sum $\pi$ ($180^\circ$) ho, toh $\cot A + \cot B = 0$ hota hai. 6 terms me 3 complementary pairs ban kar total 0 denge. Sahi uttar (a) hai.
20. sin²(π/8) + sin²(3π/8) + sin²(5π/8) + sin²(7π/8) = ?
    sin²(π/8) + sin²(3π/8) + sin²(5π/8) + sin²(7π/8) का मान क्या होगा?
    (A) (a) 1/2

    (B) (b) 2

    (C) (c) 3/2

    (D) (d) 3/4
    ✅ Answer & Explanation

    Sahi jawab: B) (b) 2

    Explanation: Method 1 (Allied Angles & Identity Method): Step 1: Supplementary angles se convert karein: - $\sin(7\pi/8) = \sin(\pi - \pi/8) = \sin(\pi/8) \implies \sin^2(7\pi/8) = \sin^2(\pi/8)$ - $\sin(5\pi/8) = \sin(\pi - 3\pi/8) = \sin(3\pi/8) \implies \sin^2(5\pi/8) = \sin^2(3\pi/8)$ Step 2: Expression ko group karein: $$\text{Sum} = 2\left[\sin^2\left(\frac{\pi}{8}\right) + \sin^2\left(\frac{3\pi}{8}\right)\right]$$ Step 3: Chuki $\frac{\pi}{8} + \frac{3\pi}{8} = \frac{4\pi}{8} = \frac{\pi}{2}$ complementary hain: $$\sin\left(\frac{3\pi}{8}\right) = \cos\left(\frac{\pi}{8}\right)$$ $$\sin^2\left(\frac{\pi}{8}\right) + \cos^2\left(\frac{\pi}{8}\right) = 1$$ Step 4: Total value evaluate karein: $$\text{Sum} = 2 \times 1 = 2$$ Correct option (b) 2 hai. Method 2 (Direct Identity Method): $\sin^2 A + \sin^2 B = 1$ jab $A + B = 90^\circ$. Do pairs banne ke karan: $1 + 1 = 2$. Sahi uttar (b) hai.
21. What is the value of (4/3) cot²(π/6) + 3 cos² 150° - 4 cosec² 45° + 8 sin(π/2) ?
    (4/3) cot²(π/6) + 3 cos² 150° - 4 cosec² 45° + 8 sin(π/2) का मान क्या होगा?
    (A) (a) 25/4

    (B) (b) 1

    (C) (c) - 7/2

    (D) (d) 13/2
    ✅ Answer & Explanation

    Sahi jawab: A) (a) 25/4

    Explanation: Method 1 (Step-by-Step Values Evaluation Method): Step 1: Sabhi individual terms ki values nikalte hain: - $\cot(\pi/6) = \cot 30^\circ = \sqrt{3} \implies \frac{4}{3}\cot^2(\pi/6) = \frac{4}{3}(\sqrt{3})^2 = \frac{4}{3} \times 3 = 4$ - $\cos 150^\circ = \cos(180^\circ - 30^\circ) = -\frac{\sqrt{3}}{2} \implies 3\cos^2 150^\circ = 3\left(-\frac{\sqrt{3}}{2}\right)^2 = 3 \times \frac{3}{4} = \frac{9}{4}$ - $\csc 45^\circ = \sqrt{2} \implies 4\csc^2 45^\circ = 4(\sqrt{2})^2 = 4 \times 2 = 8$ - $\sin(\pi/2) = 1 \implies 8\sin(\pi/2) = 8 \times 1 = 8$ Step 2: Sabhi ko combine karein: $$\text{Value} = 4 + \frac{9}{4} - 8 + 8 = 4 + \frac{9}{4} = \frac{16 + 9}{4} = \frac{25}{4}$$ Correct option (a) 25/4 hai. Method 2 (Direct Simplification): $$4 + 2.25 - 8 + 8 = 6.25 = \frac{25}{4}$$ Sahi uttar (a) hai.
22. What is the value of [4 cos(90 - A) sin³(90 + A) - 4 sin(90 + A) cos³(90 - A)] / cos[(180 + 8A)/2] ?
    [4 cos(90 - A) sin³(90 + A) - 4 sin(90 + A) cos³(90 - A)] / cos[(180 + 8A)/2] का मान क्या होगा?
    (A) (a) 1

    (B) (b) - 1

    (C) (c) 0

    (D) (d) 2
    ✅ Answer & Explanation

    Sahi jawab: B) (b) - 1

    Explanation: Method 1 (Compound Angle Identity Transformation Method): Step 1: Terms ko simplify karein: - $\cos(90^\circ - A) = \sin A$ - $\sin(90^\circ + A) = \cos A$ Step 2: Numerator ban gaya: $$4\sin A\cos^3 A - 4\cos A\sin^3 A = 4\sin A\cos A(\cos^2 A - \sin^2 A)$$ Step 3: Double angle identities apply karein ($2\sin A\cos A = \sin 2A$ aur $\cos^2 A - \sin^2 A = \cos 2A$): $$= 2(2\sin A\cos A)(\cos 2A) = 2\sin 2A\cos 2A = \sin 4A$$ Step 4: Denominator evaluate karein: $$\cos\left(\frac{180^\circ + 8A}{2}\right) = \cos(90^\circ + 4A) = -\sin 4A$$ Step 5: Numerator ko Denominator se divide karein: $$\frac{\sin 4A}{-\sin 4A} = -1$$ Correct option (b) - 1 hai. Method 2 (Value-Putting Method): Angle $A = 15^\circ$ assume karein: - Numerator $= \sin(4 \times 15^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$ - Denominator $= -\sin 60^\circ = -\frac{\sqrt{3}}{2}$ - Ratio $= -1$. Sahi uttar (b) hai.
23. What is the value of tan(π/4 + A) × tan(3π/4 + A) ?
    tan(π/4 + A) × tan(3π/4 + A) का मान क्या होगा?
    (A) (a) 0

    (B) (b) 1

    (C) (c) (cotA)/2

    (D) (d) - 1
    ✅ Answer & Explanation

    Sahi jawab: D) (d) - 1

    Explanation: Method 1 (Allied Angles & Tangent Product Method): Step 1: Dusre angle ko convert karte hain: $$\frac{3\pi}{4} + A = \pi - \left(\frac{\pi}{4} - A\right)$$ $$\tan\left(\frac{3\pi}{4} + A\right) = -\tan\left(\frac{\pi}{4} - A\right)$$ Step 2: Expression ban gaya: $$-\left[\tan\left(\frac{\pi}{4} + A\right) \times \tan\left(\frac{\pi}{4} - A\right)\right]$$ Step 3: Standard formula: $\tan(45^\circ + A) \times \tan(45^\circ - A) = 1$: $$-[1] = -1$$ Correct option (d) - 1 hai. Method 2 (Value-Putting Method): $A = 0$ assume karein: $$\tan(\pi/4) \times \tan(3\pi/4) = 1 \times (-1) = -1$$ Sahi uttar (d) hai.
24. What is the value of sin (630° + A) + cosA ?
    sin (630° + A) + cosA का मान क्या होगा?
    (A) (a) √3/2

    (B) (b) 1/2

    (C) (c) 0

    (D) (d) 2√3
    ✅ Answer & Explanation

    Sahi jawab: C) (c) 0

    Explanation: Method 1 (Large Angle Reduction Method): Step 1: 630° ko $360^\circ$ ke multiples me break karein: $$630^\circ = 360^\circ + 270^\circ$$ $$\sin(630^\circ + A) = \sin(360^\circ + 270^\circ + A) = \sin(270^\circ + A)$$ Step 2: 4th quadrant me sine negative hota hai aur axis badal kar cosine banta hai: $$\sin(270^\circ + A) = -\cos A$$ Step 3: Value substitute karein: $$-\cos A + \cos A = 0$$ Correct option (c) 0 hai. Method 2 (Value-Putting Method): $A = 0^\circ$ assume karein: $$\sin 630^\circ + \cos 0^\circ = \sin(720^\circ - 90^\circ) + 1 = -\sin 90^\circ + 1 = -1 + 1 = 0$$ Sahi uttar (c) hai.
25. What is the value of cos(90 - B) sin(C - A) + sin(90 + A) cos(B + C) - sin(90 - C) cos(A + B) ?
    cos(90 - B) sin(C - A) + sin(90 + A) cos(B + C) - sin(90 - C) cos(A + B) का मान क्या होगा?
    (A) (a) 1

    (B) (b) sin(A + B - C)

    (C) (c) cos(B + C - A)

    (D) (d) 0
    ✅ Answer & Explanation

    Sahi jawab: D) (d) 0

    Explanation: Method 1 (Allied Angles & Expansion Method): Step 1: First-step conversions: - $\cos(90^\circ - B) = \sin B$ - $\sin(90^\circ + A) = \cos A$ - $\sin(90^\circ - C) = \cos C$ Step 2: Expression ban gaya: $$\sin B\sin(C - A) + \cos A\cos(B + C) - \cos C\cos(A + B)$$ Step 3: Sabhi brackets ko standard compound formulas se expand karein: - $\sin B(\sin C\cos A - \cos C\sin A) = \sin B\sin C\cos A - \sin A\sin B\cos C$ - $\cos A(\cos B\cos C - \sin B\sin C) = \cos A\cos B\cos C - \cos A\sin B\sin C$ - $-\cos C(\cos A\cos B - \sin A\sin B) = -\cos A\cos B\cos C + \sin A\sin B\cos C$ Step 4: Sabhi terms ko add karein (sab aapas me cancel ho jate hain): $$(\sin B\sin C\cos A - \cos A\sin B\sin C) + (-\sin A\sin B\cos C + \sin A\sin B\cos C) + (\cos A\cos B\cos C - \cos A\cos B\cos C) = 0$$ Correct option (d) 0 hai. Method 2 (Value-Putting Method): $A = B = C = 0^\circ$ assume karein: $$\cos 90^\circ \sin 0^\circ + \sin 90^\circ \cos 0^\circ - \sin 90^\circ \cos 0^\circ = 0 + 1(1) - 1(1) = 0$$ Sahi uttar (d) hai.
26. If cos x = - 1/2 and π < x < 3π/2, then the value of 2 tan²x - 3 cosec²x is :
    यदि cos x = - 1/2 और π < x < 3π/2 है, तो 2 tan²x - 3 cosec²x का मान क्या होगा?
    (A) (a) 2

    (B) (b) 10

    (C) (c) 8

    (D) (d) 4
    ✅ Answer & Explanation

    Sahi jawab: A) (a) 2

    Explanation: Method 1 (Quadrant Angle Determination Method): Step 1: $\pi < x < \frac{3\pi}{2}$ third quadrant ko represent karta hai jahan $x = 180^\circ + 60^\circ = 240^\circ$ hai. Step 2: Third quadrant me $\tan x$ positive aur $\csc x$ negative hota hai: - $\tan 240^\circ = \tan(180^\circ + 60^\circ) = \sqrt{3} \implies \tan^2 x = 3$ - $\csc 240^\circ = -\csc 60^\circ = -\frac{2}{\sqrt{3}} \implies \csc^2 x = \frac{4}{3}$ Step 3: Target expression me substitute karein: $$2\tan^2 x - 3\csc^2 x = 2(3) - 3\left(\frac{4}{3}\right) = 6 - 4 = 2$$ Correct option (a) 2 hai. Method 2 (Identity Transformation Method): $$\tan^2 x = \sec^2 x - 1 = (-2)^2 - 1 = 3$$ $$\csc^2 x = 1 + \cot^2 x = 1 + \frac{1}{3} = \frac{4}{3}$$ $$2(3) - 3(4/3) = 6 - 4 = 2$$ Sahi uttar (a) hai.
27. If cos x = - √3/2 and π < x < 3π/2, then the value of 2 cot²x + 3 sec²x is :
    यदि cos x = - √3/2 और π < x < 3π/2 है, तो 2 cot²x + 3 sec²x का मान क्या होगा?
    (A) (a) 10

    (B) (b) 4

    (C) (c) 8

    (D) (d) 16
    ✅ Answer & Explanation

    Sahi jawab: A) (a) 10

    Explanation: Method 1 (Quadrant Angle Value Method): Step 1: $\pi < x < \frac{3\pi}{2}$ (3rd quadrant) me $\cos x = -\frac{\sqrt{3}}{2} \implies x = 180^\circ + 30^\circ = 210^\circ$. Step 2: Required values find karein: - $\cot 210^\circ = \cot(180^\circ + 30^\circ) = \sqrt{3} \implies \cot^2 x = 3$ - $\sec 210^\circ = -\sec 30^\circ = -\frac{2}{\sqrt{3}} \implies \sec^2 x = \frac{4}{3}$ Step 3: Expression me substitute karein: $$2\cot^2 x + 3\sec^2 x = 2(3) + 3\left(\frac{4}{3}\right) = 6 + 4 = 10$$ Correct option (a) 10 hai. Method 2 (Direct Identity Method): $$\sec^2 x = \frac{1}{\cos^2 x} = \frac{1}{3/4} = \frac{4}{3}$$ $$\cot^2 x = \frac{\cos^2 x}{1 - \cos^2 x} = \frac{3/4}{1/4} = 3$$ $$2(3) + 3(4/3) = 6 + 4 = 10$$ Sahi uttar (a) hai.
28. Find x if cos x = - 1/2.
    यदि cos x = - 1/2 है, तो x का मान ज्ञात कीजिए।
    (A) (a) 3π/2

    (B) (b) 2π/3

    (C) (c) 5π/2

    (D) (d) 4π/3
    ✅ Answer & Explanation

    Sahi jawab: B) (b) 2π/3

    Explanation: Method 1 (Principal Value Branch Method): Step 1: Cosine function ki principal value branch $[0, \pi]$ hoti hai. Step 2: $\cos x = -\frac{1}{2}$ second quadrant me lie karta hai: $$x = \pi - \frac{\pi}{3} = \frac{2\pi}{3} = 120^\circ$$ Correct option (b) 2π/3 hai. Method 2 (Option Verification Method): - (a) $\cos(3\pi/2) = \cos 270^\circ = 0$ - (b) $\cos(2\pi/3) = \cos 120^\circ = -\cos 60^\circ = -1/2$ (Direct Match) - (c) $\cos(5\pi/2) = 0$ - (d) $\cos(4\pi/3) = \cos 240^\circ = -1/2$ (Lekin standard SSC exam priority me first positive principal angle $2\pi/3$ ko mukhya uttar mana jata hai). Sahi uttar (b) hai.

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